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“Anyone who can contemplate
quantum mechanics without getting
dizzy hasn’t understood it.”
--Niels Bohr
Lecture 17, p 1
Special (Optional) Lecture
“Quantum Information”

One of the most modern applications of QM




Prof. Kwiat will give a special 214-level lecture on this
topic



quantum computing
quantum communication – cryptography, teleportation
quantum metrology
Sunday, Feb. 27
3 pm, 141 Loomis
Attendance is optional, but encouraged.
Lecture 17, p 2
Overview of the Course
Up to now:





General properties and equations of quantum mechanics
Time-independent Schrodinger’s Equation (SEQ) and eigenstates.
Time-dependent SEQ, superposition of eigenstates, time dependence.
Collapse of the wave function
Tunneling
This week:
 3 dimensions, angular momentum, electron spin, H atom
 Exclusion principle, periodic table of atoms
Next week:
 Molecules and solids, consequences of Q. M., Schrodinger’s cat
 Metals, insulators, semiconductors, superconductors, lasers, . .
Final Exam: Monday, Mar. 5
Homework 6: Due Saturday (March 3), 8 am
Lecture 16, p 3
Lecture 17:
Angular Momentum, Atomic States,
Spin, & Selection Rules
Lecture 17, p 4
Today
Schrödinger’s Equation for the Hydrogen Atom
 Radial wave functions
 Angular wave functions
Angular Momentum
 Quantization of Lz and L2
 Spin and the Pauli exclusion principle
 Stern-Gerlach experiment
Lecture 17, p 5
Summary of S-states of H-atom
The “s-states” (l=0, m=0) of the Coulomb potential have
no angular dependence. In general:
 nlm  r ,q ,f   Rnl  r Ylm q ,f 
but:
 n 00  r ,q ,f   Rn 0  r 
S-state wave functions are
spherically symmetric.
because Y00(q,f) is a constant.
|20(r,q,f)|2 :
Some s-state wave functions (radial part):
1 1
1 1
R10
f( x) 0.5
h( x)
0 0
.2
0.5
3
R20
2
d4( x)
0
000
2
r
x
4
000
4a
4 0
R30
0
.5
5
rx
10
000
10a
10 0
5
rx
10
http://www.falstad.com/qmatom/
15
15a
0
15
Lecture 17, p 6
Total Wave Function of the H-atom
We will now consider non-zero values of the other two
quantum numbers: l and m.
q
z
 nlm  r ,q ,f   Rnl  r Ylm q ,f 
 n “principal”
 l
“orbital”
 m “magnetic”
(n  1)
(0  l < n-1)
(-l  m  +l)
x
r
y
f
*
The Ylm(q,f) are known as “spherical harmonics”.
They are related to the angular momentum of the electron.
*
The constraints on l and m come from the boundary conditions one must impose on
the solutions to the Schrodinger equation. We’ll discuss them briefly.
Lecture 17, p 7
Quantized Angular Momentum
Linear momentum depends on the wavelength (k=2p/l):
p  k where  ( x )  eikx
Angular momentum depends on the tangential component of the
momentum. Therefore Lz depends on the wavelength as one moves
around a circle in the x-y plane. Therefore, a state with Lz has a similar
form:
LZ  m where  (r )  Ylm (q ,f )  e
imf
Lz
Re()
We’re ignoring
R(r) for now.
An important boundary condition:
An integer number of wavelengths must fit around the circle.
Otherwise, the wave function is not single-valued.
f
Reminder:
eimf = cos(mf) + i sin(mf)
This implies that m = 0, ±1, ±2, ±3, …
and Lz = 0, ±ħ, ±2ħ, ±3ħ, …
Angular momentum is quantized!!
http://www.falstad.com/qmatom/
Lecture 17, p 8
The l Quantum Number
The quantum number m reflects the component of angular momentum
about a given axis.
Lz  m where m  0,  1,  2, ...
In the angular wave function lmq,f
the quantum number l tells us the total angular momentum L.
L2 = Lx2 + Ly2 + Lz2 is also quantized. The possible values of L2 are:
L2  l (l  1)
2
where l  0, 1, 2, ...
Wave functions can be eigenstates of both L2 and LZ.
For spherically symmetric potentials, like H-atom, they can also be
eigenstates of E. Such states are called “orbitals”.
Summary of quantum numbers for the H-atom orbitals:
Principal quantum number:
n = 1, 2, 3, ….
Orbital quantum number:
l = 0, 1, 2, …, n-1
Orbital ‘magnetic’ quantum number: m = -l, -(l-1), … 0, … (l-1), l
Lecture 17, p 9
Angular Momentum & Uncertainty Principle
 
2
Note that L2  l (l  1) 2 not l.
Also, we describe angular momentum using only two numbers, l and m.
Q: Why can’t we specify all three components (e.g., L =(0,0,l) so that L2= l2?
A: The uncertainty principle doesn’t allow us to know that both Lx = 0
and Ly = 0 unless Lz = 0 also.
Proof by contradiction: Assume L =(0,0,l).
L  r  p, so if L points along the z-axis, both r and p lie in the x-y plane.
This means that Dz = 0 and Dpz = 0, violating the uncertainty principle.
Thus, L must have a nonzero Lx or Ly, making L2 somewhat larger.
We can’t specify all three components of the angular momentum vector.
This logic only works for L  0. L = (0,0,0) is allowed. It’s the s-state.
All physical quantities are subject to uncertainty relations,
not just position and momentum.
Lecture 17, p 10
Classical Picture of L-Quantization
e.g., l = 2
Lz
L  l( l  1 ) 2  2( 2  1 ) 2  6 
L = 6 
+2
+
0
-
-2
Lrp
The Angular Wave Function, Ylm(q,)
The angular wave function may be written: Ylm(q,f) = P(q)eimf
where P(q) are polynomial functions of cos(q) and sin(q).
To get some feeling for these angular distributions, we make
polar plots of the q-dependent part of |Ylm(q, f)| (i.e., P(q)):
l=0
l=1
1
Y0,0 
4p
Parametric Curve
1.2
1
Y1,1  sinq
Y1,0  cos q
Parametric Curve
1.2
1
q
1
q
q
0
0
0
1.2 1
1.2 1
1.2 1
1
1.2
0
x( t )
1
1
z
1.2
1.2
y
x
x
Length of the
dashed arrow
is the magnitude
of Ylm as a
function of q.
x11( t )
x10( t )
y( t )
Parametric Curve
1.2
0
z
1
1
y10( t )
1.2
0
1.2
y11( t )
1
z
1.2
y
y
x
Lecture 17, p 12
The Angular Wave Function, Ylm(q,)
z
l=2
z
Parametric Curve
2 2
q


Y2,0  3cos2 q  1
y
x20( t )
0
2
1.2 2
x
0
1 2
Y2,1  sin q cos q
z
Parametric Curve
2
y20( t )
2
2
q
x21( t )
0
x
1.2 1
Parametric Curve
1.2
1
1 1.2
Y2,2  sin q
2
y
0
y21( t )
1
q
1.2
z
+
x( t )
0
y
x
1.2 1
1
1.2
0
y( t )
1
1.2
Lecture 17, p 13
Act 1
How does the angular part of the wave function depend on the
principal quantum number, n?
a. The number of “lobes” increases as n increases.
b. As n increases, the wave function becomes
more concentrated in the xy plane.
c. No dependence.
Lecture 17, p 14
Solution
How does the angular part of the wave function depend on the
principal quantum number, n?
a. The number of “lobes” increases as n increases.
b. As n increases, the wave function becomes
more concentrated in the xy plane.
c. No dependence.
The principal quantum number describes the radial motion,
not the angular motion. Rnl(r) depends on n, but Ylm(q,f) does not.
Lecture 17, p 15
Cylindrical Symmetry
z
Why do none of the graphs display f-dependence?
(They all have cylindrical symmetry.)
y
For a given m, the f dependence of  is eimf. When we square x
it to find the probability, eimfe-imf = 1.
In order to see f dependence, we need a superposition of
different m’s.
z
For example, consider the superposition:
(l = 1, m = +1) & (l = 1, m = -1).
This will have an azimuthal wave function:
eif + e-if ≈ cos f, i.e., lobes along the x-axis:
z
Similar arguments explain how
to create the usual “d” lobes,
from l =2, m = 2 superpositions:
+
y
x
y
x
-
+
See Supplement
for more info.
Lecture 17, p 16
Why are these distributions important?
They govern the bonding and chemistry of atoms.
In particular, they determine the angles at which different atoms bond:
 the structure of molecules & solids.
Historical Labeling of Orbitals
Angular momentum quantum #
l=0
Notation from 19th century
spectroscopy
s “sharp”
l=1
p “principle”
l=2
d “diffuse”
l=3
f
“fundamental”
Lecture 17, p 17
Effect of l on Radial Wave Functions Rn,l
n=1
l=0
n=2
n=3
l=0
l=0
l=1
l=1
l=2
1: l < n (Total energy must always be
larger than rotational part.)
2: a. For fixed l, the number of
radial nodes increases with n.
b. For fixed n, the number of
radial nodes decreases with l.
(E = Trad + Trot + U(r) , i.e.,
‘radial KE’ decreases as
‘rotational KE’ increases ).
3: # radial nodes = (n-1) - l .
4: (r=0) = 0 for l  0
Do you understand why?
(i.e., a physics explanation)
The energy eigenvalues do not depend
at all on l.
En = -13.6 eV/n2
This is only true for the
Coulomb potential.
Lecture 17, p 18
Probability Density of Electrons
Let’s look at the angular momentum states of the hydrogen atom.
Probability density = Probability per unit volume = |nlm|2  Rnl2 Ylm2.
The density of dots plotted below is proportional to |nlm|2.
1s state
n,l,m = 1,0,0
2s state
2p states
2,0,0
2,1,{0,±1}
Lecture 17, p 19
Hydrogen Atom States: Summary
Key Points:
0
n=3
n: principal quantum #
n=2
l: orbital quantum #
ml:orbital magnetic quantum #
-5
Energy depends only on n
E
En 
 e 1
13.6 eV


2a0 n 2
n2
2
-10
For a given n, there are n
possible angular momentum states:
l = 0, 1, ..., n-1
n=1
-15
For a given l, there are 2l + 1
possible z-components: ml = -l, -(l -1), … 0 … (l -1), l
Therefore, a level with quantum number n has n2 degenerate states.
Lecture 17, p 20
Hydrogen Atom States: Summary
(n,l,ml)
0
n=3
n=2
-5
(3,0,0)
(3,1,-1), (3,1,0), (3,1,1)
(3,2,-2), (3,2,-1), (3,2,0), (3,2,1), (3,2,2)
(2,0,0)
(2,1,-1), (2,1,0), (2,1,1)
E
-10
n=1
(1,0,0)
-15
Lecture 17, p 21
Stern-Gerlach Experiment & Electron Spin
In 1922, Stern and Gerlach shot a beam of Ag atoms (with l = 0) through
a non-uniform magnetic field and detected them at a screen.
screen
oven
B
???
We can think of the atoms as tiny magnets (they have a magnetic
moment) being directed through the field. They are randomly oriented:
N
Strong B
S
Weak B
Lecture 17, p 22
Act 2
1. Consider a magnet in an
inhomogeneous field, as shown.
Which way will the magnet feel a force?
a.
b.
c.
d.
e.
Up
Down
Left
Right
No force
Stronger B
N
S
Weaker B
2. The magnets (i.e., atoms) leave the oven with random orientations.
What pattern do you expect on the screen?
Lecture 17, p 23
Solution
1. Consider a magnet in an
inhomogeneous field, as shown.
Which way will the magnet feel a force?
N
a.
b.
c.
d.
e.
N
Up
Down
Left
Right
No force
The N pole is in a stronger
field than the S pole, so its
upward force dominates.
S
S
Stronger B
Weaker B
N
S
2. The magnets (i.e., atoms) leave the oven with random orientations.
What pattern do you expect on the screen?
Lecture 17, p 24
Solution
1. Consider a magnet in an
inhomogeneous field, as shown. Which
way will the magnet feel a force?
a.
b.
c.
d.
e.
Up
Down
Left
Right
No force
The N pole is in a stronger
field than the S pole, so its
upward force dominates.
Stronger B
N
S
Weaker B
2. The magnets (i.e., atoms) leave the oven with random orientations.
What pattern do you expect on the screen?
oven
B
We expect a blob,
because the position
depends on the
random rotation angle.
Lecture 17, p 25
B-field off:
No splitting
B-field on:
Two peaks!
Gerlach's postcard, dated 8 February 1922, to Niels Bohr. It shows a
photograph of the beam splitting, with the message, in translation: “Attached
[is] the experimental proof of directional quantization. We congratulate [you]
on the confirmation of your theory.”
Lecture 17, p 26
Back to the Stern-Gerlach Experiment
screen
oven
B
You will analyze this
experiment in discussion.
The beam split in two! This marked the discovery of a new type of
angular momentum, with an ms quantum number that can take on only
two values:
(s = ½) ms = ±½
The new kind of angular momentum is called the electron “SPIN”. Why?
If the electron were spinning on its axis, it would have angular momentum and
a magnetic moment (because it’s charged) regardless of its spatial motion.
However, this “spinning” ball picture is not realistic, because it would require
the point-like electron to spin so fast that parts would travel faster than c!
So we can’t picture the spin in any simple way … the electron’s spin is simply
another degree-of-freedom available to electron.
Note: Most particles have spin (protons, neutrons, quarks, photons…) Lecture 17, p 27
Electron Spin
We need FOUR quantum numbers to specify the electronic
state of a hydrogen atom.
n, l, ml, ms (where ms = -½ and +½)
Actually, the nucleus (a proton) also has spin, so we must
specify its ms as well …
We’ll work some example problems next time.
Lecture 17, p 28
Next Week
Multi-electron Atoms
Covalent bonds
Electron energy bands in Solids
QM in everyday life
Lecture 17, p 29
Supplement:
Superposition and Chemical Bonding
Chemical bonds are stronger when the bonding electrons in each atom lie on the
side near the other atom. This happens as a result of superposition. A state with
definite (l,m) is symmetrical, but a superposition does not have to be. The example
here is called an “sp hybrid”:
Y1,0  cos q
1
Y0,0 
4p
Parametric Curve
1.2
1
Parametric Curve
1.2
1
q
Y00 + Y10
q
+
x10( t )
y( t )
0
0
1.2 1
1.2 1
1
1.2
0
x( t )
z
1
1
1.2
1.2
y
x
0
z
1
y10( t )
1.2
y
x
Lecture 17, p 30
Supplement: Chemistry Notation
From chemistry you may be familiar with states like dxy, etc.
How do these relate to our Ylm?
 “d” means l=2.
 “xy” stands for a particular superposition of different m’s.
dxy = (Y22+Y2 -2)/2.
z
y
The probability distribution is shown here:
Which set of states is ‘right’?
x
It depends on the problem you want to solve.
 In a strong magnetic field the “m” states are (approximately) the energy
eigenstates, because the magnetic moment determines the energy.
 In a crystalline environment, states like “xy” may be better, because the
interaction with nearby atoms dominates the energy.
Lecture 17, p 31